Finding Exact Values of Composite Functions with Inverse Trigonometric Functions
There are times when we need to compose a trigonometric function with an inverse trigonometric function. In these cases, we can usually find exact values for the resulting expressions without resorting to a calculator. Even when the input to the composite function is a variable or an expression, we can often find an expression for the output. To help sort out different cases, let f(x) and g(x) be two different trigonometric functions belonging to the set {sin(x), cos(x), tan(x)} and let [latex]f^{−1}(y)[/latex] and [latex]g^{−1}(y)[/latex] be their inverses.
Evaluating Compositions of the Form [latex]f\left(f^{−1}(y)\right)[/latex] and [latex]f^{−1}(f(x))[/latex]
For any trigonometric function, [latex]f(f^{−1}(y))=y[/latex] for all y in the proper domain for the given function. This follows from the definition of the inverse and from the fact that the range of f was defined to be identical to the domain of [latex]f^{−1}[/latex]. However, we have to be a little more careful with expressions of the form [latex]f^{−1}(f(x))[/latex].
compositions of a trigonometric function and its inverse
[latex]\begin{align} &\sin(\sin^{−1}x)=x\text{ for }−1\leq x\leq1\\ &\cos(\cos^{−1}x)=x\text{ for }−1\leq x\leq1 \\ &\tan(\tan^{−1}x)=x\text{ for }−\infty\text{ < }x\text{ < }\infty \end{align}[/latex]
[latex]\begin{align} \hfill &\sin^{−1}(\sin x)=x\text{ only for }−\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \hfill \\ &\cos^{−1}(\cos x)=x\text{ only for }0\leq x\leq\pi \hfill \\ &\tan^{−1}(\tan x)=x\text{ only for }−\frac{\pi}{2}\text{ < }x\text{ < }\frac{\pi}{2} \end{align}[/latex]
No. This equation is correct if x belongs to the restricted domain [latex]\left[−\frac{\pi}{2},\frac{\pi}{2}\right][/latex], but sine is defined for all real input values, and for x outside the restricted interval, the equation is not correct because its inverse always returns a value in [latex]\left[−\frac{\pi}{2},\frac{\pi}{2}\right][/latex]. The situation is similar for cosine and tangent and their inverses. For example, [latex]\sin^{−1}\left(\sin\left(\frac{3\pi}{4}\right)\right)=\frac{\pi}{4}[/latex].
- If θ is in the restricted domain of f, then [latex]f^{−1}(f(\theta))=\theta[/latex].
- If not, then find an angle ϕ within the restricted domain of f such that [latex]f(\phi)=f(\theta)[/latex]. Then [latex]f^{−1}(f(\theta))=\phi[/latex].
- [latex]\sin^{−1}(\sin(\frac{\pi}{3}))[/latex]
- [latex]\sin^{−1}(\sin(\frac{2\pi}{3}))[/latex]
- [latex]\cos^{−1}(\cos(\frac{2\pi}{3}))[/latex]
- [latex]\cos^{−1}(\cos(−\frac{\pi}{3}))[/latex]
Evaluate [latex]\tan^{−1}\left(\tan\left(\frac{\pi}{8}\right)\right)[/latex] and [latex]\tan^{−1}\left(\tan\left(\frac{11\pi}{9}\right)\right)[/latex].
Evaluating Compositions of the Form [latex]f^{−1}(g(x))[/latex]
Now that we can compose a trigonometric function with its inverse, we can explore how to evaluate a composition of a trigonometric function and the inverse of another trigonometric function. We will begin with compositions of the form [latex]f^{−1}(g(x))[/latex]. For special values of x, we can exactly evaluate the inner function and then the outer, inverse function. However, we can find a more general approach by considering the relation between the two acute angles of a right triangle where one is θ, making the other [latex]\frac{\pi}{2}−\theta[/latex]. Consider the sine and cosine of each angle of the right triangle.

Because [latex]\cos\theta=\frac{b}{c}=\sin\left(\frac{\pi}{2}−\theta\right)[/latex], we have [latex]\sin^{−1}(\cos\theta)=\frac{\pi}{2}−\theta\text{ if }0\leq\theta\leq\pi[/latex]. If θ is not in this domain, then we need to find another angle that has the same cosine as θ and does belong to the restricted domain; we then subtract this angle from [latex]\frac{\pi}{2}[/latex]. Similarly, [latex]\sin\theta=\frac{a}{c}=\cos\left(\frac{\pi}{2}−\theta\right)[/latex], so [latex]\cos^{−1}(\sin\theta)=\frac{\pi}{2}−\theta\text{ if }−\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}[/latex]. These are just the function-cofunction relationships presented in another way.
- If x is in [0,π], then [latex]\sin^{−1}(\cos x)=\frac{\pi}{2}−x[/latex].
- If x is not in [0,π], then find another angle y in [0,π] such that [latex]\cos y=\cos x[/latex].
[latex]\sin^{−1}(\cos x)=\frac{\pi}{2}−y[/latex]
- If x is in [latex]\left[−\frac{\pi}{2},\frac{\pi}{2}\right][/latex], then [latex]\cos^{−1}(\sin x)=\frac{\pi}{2}−x[/latex].
- If x is not in [latex]\left[−\frac{\pi}{2},\frac{\pi}{2}\right][/latex], then find another angle y in [latex]\left[−\frac{\pi}{2}\text{, }\frac{\pi}{2}\right][/latex] such that [latex]\sin y=\sin x[/latex].
[latex]\cos^{−1}(\sin x)=\frac{\pi}{2}−y[/latex]
- by direct evaluation.
- by the method described previously.
Evaluating Compositions of the Form [latex]f(g^{−1}(x))[/latex]
To evaluate compositions of the form [latex]f(g^{−1}(x))[/latex], where f and g are any two of the functions sine, cosine, or tangent and x is any input in the domain of [latex]g−1[/latex], we have exact formulas, such as [latex]\sin\left({\cos}^{−1}x\right)=\sqrt{1−{x}^{2}}[/latex]. When we need to use them, we can derive these formulas by using the trigonometric relations between the angles and sides of a right triangle, together with the use of Pythagoras’s relation between the lengths of the sides. We can use the Pythagorean identity, [latex]\sin^{2}x+cos^{2}x=1[/latex], to solve for one when given the other. We can also use the inverse trigonometric functions to find compositions involving algebraic expressions.
Example 8: Evaluating the Composition of a Sine with an Inverse Tangent
Find an exact value for [latex]\sin\left(\tan^{−1}\left(\frac{7}{4}\right)\right)[/latex].
Evaluate [latex]\cos(\sin^{−1}(\frac{7}{9}))[/latex].

